Question

The sum of the first $n$ terms of an A.P. is $4n^2+2n$. Find the $n^{th}$ term of this A.P.

Answer 1

$S_n=4n^2+2n$

$S_{n-1}=4{(n-1)}^2+2(n-1)$

$T_n=S_n-S_{n-1}$

$\Rightarrow T_n=(4n^2+2n)-(4{(n-1)}^2+2(n-1))$

$=4(n^2+{(n-1)}^2)+2(n-n+1)$

$=4\left((n-n+1)(n+n-1)\right)+2$ on expanding

$=4(2n-1)+2$

$=8n-4+2=8n-2$